= Solution
Because $\Delta K=0$, the $(\widetilde C-\widetilde B)K^2$ terms cancel between the initial and final states. The two positive-shift branches are consequently
$$
\boxed{\begin{aligned}
\Delta J=+1:&\quad\widetilde\nu_{\rm S}=\widetilde\nu_L-2\widetilde B(J+1),\\
\Delta J=+2:&\quad\widetilde\nu_{\rm S}=\widetilde\nu_L-(4J+6)\widetilde B.
\end{aligned}}
$$
The $+2$ branch gives shifts $6\widetilde B,10\widetilde B,14\widetilde B,\ldots$, as for the diatomic. The $+1$ branch fills in a series spaced by $2\widetilde B$, so the combined nonzero series is \b[$4\widetilde B,6\widetilde B,8\widetilde B,10\widetilde B,\ldots$].
There is a boundary qualification to the abbreviated rules printed in the question: for an ordinary symmetric <polarizability tensor>, $\Delta J=\pm1$ has zero strength when $K=0$. Thus $J=0,K=0\to J=1,K=0$ does not produce a $2\widetilde B$ line; the first $+1$ transition can start at $J=1,|K|=1$ and has shift $4\widetilde B$. This follows from the rank-two angular-momentum coupling: the all-zero-projection coupling for $J$ and $J+1$ vanishes because $J+2+(J+1)$ is odd. If one applies only the printed inequalities without this matrix-element restriction, $2\widetilde B$ is a formal position, but its intensity is zero.
The lower panel of the figure in part (a)(ii) marks the two branches and their overlaps. \b[The useful spacing is $2\widetilde B$, not the diatomic value $4\widetilde B$.]
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